Question

A ball is projected upward from the ground with a speed of 50 m/s.

Answer 1

Answer:

(a) Initial velocity of ball (u)=50m/s, acceleration of ball =−g,

final velocity at the highest point (v)=0

So applying the 3rd equation of motion we get:

v  

2

=u  

2

−2gh  

max

​  

 

⇒0=50  

2

−2×10×h  

max

​  

 

⇒h  

max

​  

=  

20

2500

​  

=125m

(b) Let the time required to reach max height be t. Then applying 1st equation of motion we get:

v=u−gt

⇒0=50−10t

⇒t=5s

(c) Let speed at half of max height be V then:

V  

2

=50  

2

−2g  

2

125

​  

 

⇒V  

2

=2500−1250=1250

⇒V=  

1250

​  

=35.35m/s.

Step-by-step explanation: